The chords AB and CD of the circle meet at point E. AE: EB = 1: 3, CD = 20, DE = 5. Find AB.

Let the length of the segment AE = X cm, then the length of the segment BE = 3 * X cm.

The length of the segment is CE = CD – DE = 20 – 5 = 15 cm.

By the property of intersecting chords, the product of the lengths of a segment of one chord is equal to the product of dyn segments of the other chord.

AE * BE = CE * DE.

X * 3 * X = 15 * 5.

3 * X ^ 2 = 75.

X ^ 2 = 75/3 = 25.

X = AE = 5 cm, then BE = 3 * 5 = 15 cm.

AB = AE + BE = 5 + 15 = 20 cm.

Answer: The length of the chord AB is 20 cm.